Question

3. Check whether 6" can end with the digit 0 for any natural number n.

4. Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 +5 are compositenumbers​

Answer 1

Answer:

no

Step-by-step explanation:

It can be observed that 5 is not in the prime factoraisation of 6n . hence, for the any value of n , 6n will not be divisible by 5. So 6n cannot end with the digit 0 for any natural number n

Answer 2

3. Let n = 1,2 and 3

so,

${6}^{n} = {6}^{1} = 6 \\ \: \: \: \: \: \: \: \: \: \: \: \: {6}^{2} = 36 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: {6}^{3} = 216$

Here, we can see that every example is end with the digit 6.

No, 6ⁿ can never end with the digit 0 for any natural number n.

4. 7 x 11 x 13 + 13

= 13 (7 x 11 + 1)

= 13 ( 77 + 1)

= 13 (78)

= 13 x 2 x 3 x 13

it is the product of prime factors 2, 3 and 13 so it is a composite number.

And,

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

= 5 ( 7 x 6 x 4 x 3 x 2 x 1 + 1)

= 5 ( 1008 + 1)

= 5 ( 1009)

it is the product of prime factors 5 and 1009 so it is also a composite number.

Hope it helps!!!

please give me thanks I need it please give me so many thanks!!!